The acidity of a water with initial pH of 5.5 was tested using the following procedure:
The pH of a volume of 50.00 mL water sample was adjusted to 4.0 using 5 mL 0.1 N H2SO4 solution. Then this sample was titrated using 0.2 N NaOH. After addition of 7 mL of 0.2 N NaOH, the pH of this water reached to 8.2. What is the acidity of the water sample express as CaCO3 in mg/L?
To determine the acidity of the water sample expressed as CaCO3 in mg/L, we need to calculate the amount of CaCO3 required to neutralize the acid present in the water.
First, let's determine the number of moles of H2SO4 added to adjust the pH.
We know that the volume of 0.1 N H2SO4 added is 5 mL (which is equivalent to 0.005 L). The concentration of the H2SO4 solution is 0.1 N, which means it contains 0.1 moles of H2SO4 per liter.
Number of moles of H2SO4 = concentration * volume = 0.1 N * 0.005 L = 0.0005 moles.
Since one mole of H2SO4 can provide two moles of H⁺ ions, the number of moles of H⁺ ions added to the water is 2 * 0.0005 = 0.001 moles.
Next, let's determine the number of moles of NaOH added during the titration.
The volume of 0.2 N NaOH added is 7 mL (which is equivalent to 0.007 L). The concentration of the NaOH solution is 0.2 N, which means it contains 0.2 moles of NaOH per liter.
Number of moles of NaOH = concentration * volume = 0.2 N * 0.007 L = 0.0014 moles.
During the titration, each mole of NaOH reacts with one mole of H⁺ ions. Since we added 0.0014 moles of NaOH, it means that 0.0014 moles of H⁺ ions were neutralized.
So, the remaining number of moles of H⁺ ions in the water sample after titration is 0.001 - 0.0014 = -0.0004 moles. Note that the negative sign indicates that the water has a basic pH after titration.
To convert the moles of H⁺ ions to mg/L of CaCO3, we need to use the formula weight of CaCO3, which is 100.1 g/mol.
Number of moles of CaCO3 = |Number of moles of H⁺ ions| / 2.
Number of moles of CaCO3 = 0.0004 / 2 = 0.0002 moles.
To convert moles to milligrams, we multiply by the molar mass and then by 1000 to convert grams to milligrams.
Mass of CaCO3 = Number of moles of CaCO3 * molar mass of CaCO3 * 1000.
Mass of CaCO3 = 0.0002 moles * 100.1 g/mol * 1000 = 20 mg.
Therefore, the acidity of the water sample expressed as CaCO3 is 20 mg/L.